Separable trace-class determinant theorem recorded externally

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a separable complex Hilbert space, including the zero space, and let $A\in {\mathcal{S}}_1(K)$ (Trace class operator). For every nonzero eigenvalue $\lambda$ of $A$, its algebraic multiplicity is

where the increasing generalized kernels stabilize and this dimension is finite. List all nonzero eigenvalues $({\lambda }_j(A))$ with those multiplicities; the list is finite or countable and may be empty. With $(s_j(A))$ the singular values (Absolute value and singular values of a compact operator), the following results are recorded externally.

1. The eigenvalues are absolutely summable and ${\sum }_j\mathrm{\mid }{\lambda }_j(A)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_1\mathrm{.}$
2. There is an entire function $D_A$ such that, locally uniformly in $z$, $D_A(z)={\prod }_j(1+z{\lambda }_j(A))\mathrm{.}$ The empty product is $1$ and the empty eigenvalue sum is $0$.
3. If finite-rank $A_n$ satisfy $\mathrm{\parallel }{A}_n-A{\mathrm{\parallel }}_1\to 0$, then $\det (I+z{A}_n)\to D_A(z)$ locally uniformly. For finite-rank $F$, this is the ordinary determinant of $(I+zF){\mathrm{\mid }}_E$ for any finite-dimensional subspace $E$ containing $\mathrm{ran}F$; it is independent of $E$.
4. One has $D_A(0)=1,D_A^{\mathrm{\prime }}(0)={\mathrm{tr}}_K(A),\mathrm{\mid }{D}_A(z)\mathrm{\mid }\le {\prod }_j(1+\mathrm{\mid }z\mathrm{\mid }{s}_j(A))\le e^{\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }A{\mathrm{\parallel }}_1}\mathrm{.}$ For every $\epsilon >0$ there is $C_{\epsilon }$ with $\mathrm{\mid }{D}_A(z)\mathrm{\mid }\le C_{\epsilon }{e}^{\epsilon \mathrm{\mid }z\mathrm{\mid }}$.
5. For $A,B\in {\mathcal{S}}_1(K)$, $\mathrm{\mid }{D}_A(z)-D_B(z)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }A-B{\mathrm{\parallel }}_1{e}^{1+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }A{\mathrm{\parallel }}_1+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }B{\mathrm{\parallel }}_1},$ and $D_{A+B+AB}(1)=D_A(1)D_B(1)$.
6. The value $D_A(z)$ vanishes exactly when $I+zA$ is not boundedly invertible. If $\lambda \mathrm{\ne }0$ is an eigenvalue, then $-1\mathrm{/}\lambda$ is a zero of order equal to its algebraic multiplicity.

These assertions include the zero-space conventions: its unique operator has determinant identically $1$, trace $0$, and an empty eigenvalue list.

This is a source-backed external theorem, not a local exterior-power or Hadamard-factorization proof. In the cited proof of the zero criterion, the complementary factor is $I+z(I-P_{\lambda })A$; a printed omission of $A$ in one sentence is not copied here.